Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 6

For the reaction $$5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}$$ it was found that at a particular instant bromine was being formed at the rate of \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). At that instant, at what rate is (a) water being formed? (b) bromide ion being oxidized? (c) \(\mathrm{H}^{+}\) being consumed?

Short Answer

Expert verified
Question: At a particular moment in a reaction, the rate of bromine formation is 0.039 mol/L⋅s. Determine the rates of formation or consumption of water, bromide ion, and hydrogen ion at that instant. Answer: At that particular instant, water is being formed at a rate of 0.039 mol/L⋅s, bromide ions are being oxidized at a rate of 0.065 mol/L⋅s, and hydrogen ions are being consumed at a rate of 0.078 mol/L⋅s.

Step by step solution

01

a) Rate of water formation

To determine the rate of water formation, we need to compare the stoichiometric coefficients of water and bromine in the balanced equation. We can see that the coefficients are 3 and 3, respectively. Since the rate of bromine formation has been given as \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), the rate of water formation would be the same since their coefficients are equal. Therefore, the rate of water formation is: $$0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}.$$
02

b) Rate of bromide ion oxidation

To find the rate at which bromide ions are being oxidized, we need to compare the stoichiometric coefficients of bromide ions and bromine in the balanced equation. We can see that they are 5 and 3, respectively. To find the rate of bromide ion oxidation, we can set up a proportion: $$\frac{\text{Rate of Br}^{-}}{\text{Rate of Br}_{2}} = \frac{5}{3}$$ Therefore, the rate of bromide ion oxidation is: $$\text{Rate of Br}^{-} = \frac{5}{3} \times 0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} = 0.065 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}.$$
03

c) Rate of H+ ion consumption

To determine the rate at which H+ ions are being consumed, we need to compare the stoichiometric coefficients of H+ ions and bromine in the balanced equation. We can see that they are 6 and 3, respectively. To find the rate of H+ ion consumption, we can set up a proportion: $$\frac{\text{Rate of H}^{+}}{\text{Rate of Br}_{2}} = \frac{6}{3}$$ Therefore, the rate of H+ ion consumption is: $$\text{Rate of H}^{+} = 2 \times 0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} = 0.078 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}.$$ In summary, at that particular instant, water is being formed at a rate of \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), bromide ions are being oxidized at a rate of \(0.065 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), and H+ ions are being consumed at a rate of \(0.078 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The gas-phase reaction between hydrogen and iodine $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$$ proceeds with a rate constant for the forward reaction at \(700^{\circ} \mathrm{C}\) of \(138 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) and an activation energy of \(165 \mathrm{~kJ} / \mathrm{mol}\). (a) Calculate the activation energy of the reverse reaction given that \(\Delta H_{i}^{\circ}\) for HI is \(26.48 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta H_{i}^{\circ}\) for \(\mathrm{I}_{2}(\mathrm{~g})\) is \(62.44 \mathrm{~kJ} / \mathrm{mol}\). (b) Calculate the rate constant for the reverse reaction at \(700^{\circ} \mathrm{C}\). (Assume \(\mathrm{A}\) in the equation \(k=\mathrm{Ae}^{-\mathcal{P}_{2} / \mathrm{RT}^{\prime}}\) is the same for both forward and reverse reactions.) (c) Calculate the rate of the reverse reaction if the concentration of HI is \(0.200 M\). The reverse reaction is second-order in HI.

In a first-order reaction, suppose that a quantity \(X\) of a reactant is added at regular intervals of time, \(\Delta t\). At first the amount of reactant in the system builds up; eventually, however, it levels off at a saturation value given by the expression $$\text { saturation value }=\frac{X}{1-10^{-a}} \quad \text { where } a=0.30 \frac{\Delta t}{t_{1 / 2}}$$ This analysis applies to prescription drugs, of which you take a certain amount each day. Suppose that you take \(0.100 \mathrm{~g}\) of a drug three times a day and that the half-life for elimination is \(2.0\) days. Using this equation, calculate the mass of the drug in the body at saturation. Suppose further that side effects show up when \(0.500 \mathrm{~g}\) of the drug accumulates in the body. As a pharmacist, what is the maximum dosage you could assign to a patient for an 8 -h period without causing side effects?

Consider the following hypothetical reaction: $$\mathrm{X}(g) \longrightarrow \mathrm{Y}(g)$$ A 200.0-mL flask is filled with \(0.120\) moles of \(\mathrm{X}\). The disappearance of \(\mathrm{X}\) is monitored at timed intervals. Assume that temperature and volume are kept constant. The data obtained are shown in the table below. $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { moles of } \mathrm{X} & 0.120 & 0.103 & 0.085 & 0.071 & 0.066 \\ \hline\end{array}$$ (a) Make a similar table for the appearance of Y. (b) Calculate the average disappearance of \(\mathrm{X}\) in \(\mathrm{M} / \mathrm{s}\) in the first two twenty-minute intervals. (c) What is the average rate of appearance of \(\mathrm{Y}\) between the 20 - and 60-minute intervals?

The decomposition of nitrosyl chloride $$\mathrm{NOCl}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g)$$ is a second-order reaction. If it takes \(0.20 \mathrm{~min}\) to decompose \(15 \%\) of a \(0.300 \mathrm{M}\) solution of nitrosyl chloride, what is \(k\) for the reaction?

The following reaction is second-order in A and first-order in B. $$\mathrm{A}+\mathrm{B} \longrightarrow \text { products }$$ (a) Write the rate expression. (b) Consider the following one-liter vessel in which each square represents a mole of \(\mathrm{A}\) and each circle represents a mole of \(\mathrm{B}\). What is the rate of the reaction in terms of \(k ?\) (c) Assuming the same rate and \(k\) as (b), fill the similar one-liter vessel shown in the figure with an appropriate number of circles (representing B).
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free